I have written this code for delhi's even odd. But it says to think much efficiently. can you help me by giving some clues

#1

#include
using namespace std;
int main() {
int N;
int no;
cin>>N;
while(N>0) {
cin>>no;
int s=0;
while(no>=0) {
int p = no%10;
s=s+p;
no=no/10;
}
((s%2==0 && s%4==0) || (s%2!=0 && s%3==0))? cout<<“YES” : cout<<“NO”;
N=N-1;
}
return 0;
}

#2

it is mentioned that if the sum of digits which are even is divisible by 4 or sum of digits which are odd in that number is divisible by 3. which means you need to maintain 2 sum odd_sum and even_sum and add even digits in even_sum, odd digits in odd_sum and check the condition
Code

#3

@tarunasree123 You reopened the doubt. Is your doubt resolved?

#4

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